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Suppose that we have following quadratic equation containing some constant $a$ $$ax^2-(1-2a)x+a-2=0.$$ We have to find all integers $a$,for which this equation has rational roots.

First I have tried to determine for which $a$ this equation has a real solution, so I calculated the discriminant (also I guessed that $a$ must not be equal to zero, because in this situation it would be a linear form, namely $-x-2=0$ with the trivial solution $x=-2$).

So $$D=(1-2a)^2-4a(a-2)$$ if we simplify,we get $$D=1-4a+4a^2-4a^2+8a=1+4a$$

So we have the roots: $x_1=(1-2a)-\frac{\sqrt{1+4a})}{2a}$ and $x_2=(1-2a)+\frac{\sqrt{1+4a})}{2a}$

Because we are interesting in rational numbers, it is clear that $\sqrt{1+4a}$ must be a perfect square, because $a=0$ I think is not included in solution set,then what i have tried is use $a=2$,$a=6$,$a=12$,but are they the only solutions which I need or how can I find all values of $a$?

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I have edited both the formulas and the spelling. Sometimes I changed some words when I thought it made more sense. Please check whether everything is ok with you. –  Simon Markett Jun 26 '12 at 9:52

2 Answers 2

up vote 3 down vote accepted

Edited in response to Simon's comment.

Take any odd number; you can write it as $2m+1$, for some integer $m$; its square is $4m^2+4m+1$; so if $a=m^2+m$ then, no matter what $m$ is, you get rational roots.

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if $m=1$ then $a=8$ but $1+4*8=33$ but square root of 33 is not rational –  dato datuashvili Jun 26 '12 at 9:35
    
so maybe you mean $a=4*m^2+4*m+1$? –  dato datuashvili Jun 26 '12 at 9:39
    
There is just one 4 to many. In the question you have $4a$ and the answer is for $a$ instead. So $a=m^2+m$ –  Simon Markett Jun 26 '12 at 9:40
    
yes i see thanks –  dato datuashvili Jun 26 '12 at 9:41

put $4a+1=m^2$ for some $m$, then $4a=(m+1)(m-1)$.Now, $m$ must be odd otherwise RHS won't be divisible by 4.Let $m=2k+1$,it gives $4a+1=4k^2+4k+1 \implies a=k(k+1)$. So Put $k=1,2,3,$ and so on, you will get the required values of $a$.

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